If $x \veebar y = 5y+2$ and $x \triangleright y = xy+3x-y$, find $(2 \triangleright -4) \veebar 2$.
We don't need to find $2 \triangleright -4$ because $x \veebar y$ depends only on the right operand. Find $x \veebar 2$ $ x \veebar 2 = (5)(2)+2$ $ \hphantom{x \veebar 2} = 12$.